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A068069
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a(n) = least k which is the start of n consecutive integers each with a different number, 1 through n, of distinct prime factors.
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3
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OFFSET
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0,2
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COMMENTS
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a(n) >= n!. If the canonical factorization of k is the product of p^e(p) over primes, then the number of distinct number of prime factors is simply the count of the number of p's.
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LINKS
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Table of n, a(n) for n=0..9.
P. Erdős, Megjegyzések a Matematikai Lapok két problémájához (Remarks on two problems, in Hungarian), Mat. Lapok 11 (1960), pp. 26-32.
J.-M. De Koninck, J. B. Friedlander, and F. Luca, On strings of consecutive integers with a distinct number of prime factors, Proc. Amer. Math. Soc., 137 (2009), 1585-1592.
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FORMULA
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Koninck, Friedlander, & Luca prove that a(n) > exp(2n + o(n)), but note that an earlier result of Erdős is "essentially equivalent". - Charles R Greathouse IV, Feb 04 2013
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EXAMPLE
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a(1) = 2 because 2 has the single prime factor 2; a(2) = 5 because 5 = 5^1 & 6 = 2*3 which have 1 & 2 prime factors respectively; a(3) = 28 because 28 = 2^2*7^1, 29 = 29^1 & 30 = 2*3*5 which have 2, 1 & 3 prime factors respectively; a(4) = 417 because 417 = 3*139, 418 = 2*11*19, 419 = 419^1 & 420 = 2^2*3*5*7 which have 2, 3, 1 & 4 prime factors (distinct) respectively and this represents a record-breaking number.
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MATHEMATICA
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k = 3; Do[k = k - n; a = Table[ Length[ FactorInteger[i]], {i, k, k + n - 1}]; b = Table[i, {i, 1, n}]; While[ Sort[a] != b, k++; a = Drop[a, 1]; a = Append[a, Length[ FactorInteger[k]]]]; Print[k - n + 1], {n, 1, 7}]
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CROSSREFS
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Cf. A067665.
Sequence in context: A324264 A138293 A316972 * A292499 A306893 A105787
Adjacent sequences: A068066 A068067 A068068 * A068070 A068071 A068072
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KEYWORD
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more,nonn
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AUTHOR
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Robert G. Wilson v, Feb 20 2002
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EXTENSIONS
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One more term from Labos Elemer, May 26 2003
One more term from Donovan Johnson, Apr 03 2008
Corrected example and a(9) from Donovan Johnson, Aug 31 2010
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STATUS
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approved
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